Assume for the moment that the Fourier series works. To find
the Fourier coefficients, we note the following
orthogonality properties of sinusoids.
∀k,l:∫0Tsin2πktTcos2πltTdt=0
k
l
t
0
T
2
k
t
T
2
l
t
T
0
(1)
∫0Tsin2πktTsin2πltTdt=T2ifk=l∧k≠0∧l≠00ifk≠l∨k=0=l
t
0
T
2
k
t
T
2
l
t
T
T
2
k
l
k
0
l
0
0
k
l
k
0
l
(2)
∫0Tcos2πktTcos2πltTdt=T2ifk=1∧k≠0∧l≠0Tifk=0=l0ifk≠l
t
0
T
2
k
t
T
2
l
t
T
T
2
k
1
k
0
l
0
T
k
0
l
0
k
l
(3)
To use these, let's multiply the Fourier series for a square wave (
st=
a
0
+∑k=1∞
a
k
cos2πktT+∑k=1∞
b
k
sin2πktT
s
t
a
0
k
1
a
k
2
k
t
T
k
1
b
k
2
k
t
T
) by
cos2πltT
2
l
t
T
and integrate. The idea is that, because integration is linear,
the integration will sift out all but the term involving
a
l
a
l
.
∫0Tstcos2πltTdt=∫0T
a
0
cos2πltTdt+∑k=1∞
a
k
∫0Tcos2πktTcos2πltTdt+∑k=1∞
b
k
∫0Tsin2πktTcos2πltTdt
t
0
T
s
t
2
l
t
T
t
0
T
a
0
2
l
t
T
k
1
a
k
t
0
T
2
k
t
T
2
l
t
T
k
1
b
k
t
0
T
2
k
t
T
2
l
t
T
(4)
The first and third terms are zero; in the second, the only
non-zero term in the sum results when the indices
kk and
ll are equal (but not zero), in
which case we obtain
a
l
T2
a
l
T
2
. If
k=0=l
k
0
l
, we obtain
a
0
T
a
0
T
. Consequently,
∀l,l≠0:
a
l
=2T∫0Tstcos2πltTdt
l
l
0
a
l
2
T
t
0
T
s
t
2
l
t
T
All of the Fourier coefficients can be found similarly.
a
0
=1T∫0Tstdt
a
0
1
T
t
0
T
s
t
(5)
∀k,k≠0:
a
k
=2T∫0Tstcos2πktTdt
k
k
0
a
k
2
T
t
0
T
s
t
2
k
t
T
(6)
b
k
=2T∫0Tstsin2πktTdt
b
k
2
T
t
0
T
s
t
2
k
t
T
(7)
The expression for
a
0
a
0
is referred to as the average value of
st
s
t
. Why?
The average of a set of numbers is the sum divided by the
number of terms. Viewing signal integration as the limit of
a Riemann sum, the integral corresponds to the average.